[2,3] 这道题给了我们一个数组,数组中的数字可能出现一次或两次,让我们找出所有出现两次的数字,由于之前做过一道类似的题目Find the Duplicate Number,所以不是完全无从下手。这类问题的一个重要条件就是1 ≤ a[i] ≤ n (n = size of array),不然很难在O(1)空间和O(n)时间内完成。首先来看一种正负...
类似LeetCode 442. Find All Duplicates in an Array,由于元素是1~n,因此每个元素的值-1(映射到0~n-1)就可以直接当做下标。 classSolution {public:intfindDuplicate(vector<int>&nums) {for(inti=0;i<nums.size();++i){intindex=abs(nums[i])-1;if(nums[index]>0) nums[index]*=-1;elsereturnind...
Note: You must not modify the array (assume the array is read only). You must use only constant, O(1) extra space. Your runtime complexity should be less than O(n2). There is only one duplicate number in the array, but it could be repeated more than once. 哈希表法 复杂度 时间O(...
There is only one duplicate number in the array, but it could be repeated more than once. 哈希表法 复杂度 时间O(N) 空间 O(N) 思路 遍历数组时,用一个集合记录已经遍历过的数,如果集合中已经有了说明是重复。但这样要空间,不符合。 暴力法 复杂度 时间O(N^2) 空间 O(1) 思路 如果不用空间...
There is only one duplicate number in the array, but it could be repeated more than once 非常好的题目,开始是用二分做的,比如取数组为{1,2,3,3,4,5},mid应该是(5+1)/2 = 3,那么,如果小于等于mid的数的个数如果超过了3,那么重复的数字一定出现在l,mid之间,否则出现在mid + 1,r之间。以该...
Learn how to find duplicate values in a JavaScript array with this comprehensive guide, including examples and step-by-step instructions.
. There is only one duplicate number in the array, but it could be repeated more than once. 思路: 不能修改数组所以不能排序,不能用额外的空间所以不能用HashMap,考虑鸽巢原理。数的范围在1~n之间,有n+1个数 猜测重复数为n/2,计算 数组中大于n/2的次数count,如果重复的数在n/2~n之间的话,coun...
Since each value in a Set has to be unique, passing any duplicate item will be removed automatically:const numbers = [1, 2, 3, 2, 4, 5, 5, 6]; const unique = Array.from(new Set(numbers)); console.log(unique); // [ 1, 2, 3, 4, 5, 6 ] ...
- There is only one duplicate number in the array, but it could be repeated more than once. 代码如下: public int findDuplicate(int[] nums) { if (nums.length > 1){ int slow = nums[0]; int fast = nums[nums[0]]; //以下为求两个指针第一次相遇的点 while (slow != fast){ slow...
publicintfindDuplicate(int[]nums){Arrays.sort(nums);for(inti=0;i<nums.length-1;i++){if(nums[i]==nums[i+1]){returnnums[i];}}return-1;} 解法二 HashSet 判断重复数字,可以用HashSet,这个方法经常用了。 publicintfindDuplicate(int[]nums){HashSet<Integer>set=newHashSet<>();for(inti=0...